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NOTE: I've edited the question to be more focused and precise.

SETUP: Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. The spirit of this post is to inquire whether or not $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is a genuine improvement over $\operatorname{li}(x)$ in approximating $\pi(x)$ for large $x$.

Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$, and one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive).

QUESTION: Let $\Theta \in [1/2,1]$ denote the supremum of the real parts of the zeros of the Riemann zeta function. The infimum of all $t > 0$ such that $E(x) = o(x^t)$ is $\Theta$. Likewise, the infimum of all $t > 0$ such that $H(x) = o(x^t)$ is also $\Theta$. Clearly, then, the infimum of all $t > 0$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = o(x^t)$ is less than or equal to $\Theta$. Does equality hold? If not, that answers my question fully. If so, let's assume that equality holds. Let $\Theta_1$ be the infimum of all $t \in \mathbb{R}$ such that $E(x) = o(x^\Theta (\log x)^t)$. Then $\Theta_1$ is also the infimum of all $t \in \mathbb{R}$ such that $H(x) = o(x^\Theta (\log x)^t)$. Thus the infimum all $t \in \mathbb{R}$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is $o(x^\Theta (\log x)^t)$ is less than or equal to $\Theta_1$. Does equality hold?

MOTIVATION: Note that $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$. (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li(y)}}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ were smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth (in the sense asked in the QUESTION above), then that would be interesting. It would show that the product $p_{\operatorname{li(x)}}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.

NOTE: I've edited the question one last time, to be much simpler, in the hopes of getting more responses.

SETUP: Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

QUESTION: Does there exist a $t \in \mathbb{R}$ such that $ \pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) \neq O(\sqrt{x} (\log x)^t)$?

MOTIVATION:

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$, and one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive). Let $\Theta$ denote the supremum of the zeros of the Riemann zeta function. Note that both $E(x)$ and $H(x)$ are $O(x^\Theta \log x)$, and neither is $O(\sqrt{x}/\log x)$.

Note that $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$. (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li}(y)}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If the answer to the QUESTION above is NO, then $p_{\operatorname{li}(x)}\pi(x) - x \operatorname{li}(x)$ is smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth, which would imply that the product $p_{\operatorname{li}(x)}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.

Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. The spirit of this post is to inquire whether or not $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is a genuine improvement over $\operatorname{li}(x)$ in approximating $\pi(x)$ for large $x$.

Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where the functions $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$ have the same "order of growth," while one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive). Thus, the "order of growth" of $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is no larger than that of $\pi(x)-\operatorname{li}(x)$. My question is, is the opposite also true, and thus $\pi(x)-\operatorname{li}(x)$ and $\pi(x) -\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ have the "same order of growth"?

By the "same order of growth" here, I don't mean in terms of $\asymp$ or $\sim$, I mean being $o$ of the same, or almost all of the same, logarithmico-exponential functions (in Hardy's sense).
EDIT: It would take a while to explain exactly what I mean here by "same order of growth," so let me ask a narrower but more precise question.

PRECISE QUESTION: Let $\Theta \in [1/2,1]$ denote the supremum of the real parts of the zeros of the Riemann zeta function. The infimum of all $t > 0$ such that $\pi(x)-\operatorname{li}(x) = o(x^t)$ is $\Theta$. Likewise, the infimum of all $t > 0$ such that $\frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x) = o(x^t)$ is also $\Theta$. Clearly, then, the infimum of all $t > 0$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = o(x^t)$ is less than or equal to $\Theta$. Does equality hold? If so, then consider the infimum of all $t > 0$ such that the three respective functions is $o(x^\Theta (\log x)^t)$, and ask the same questions.

EDIT: Someone asks, what's the point of asking this question? One has $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$. (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li(y)}}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ were smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth, then that would be interesting. It would show that the product $p_{\operatorname{li(x)}}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.

NOTE: I've edited the question to be more focused and precise.

SETUP: Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. The spirit of this post is to inquire whether or not $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is a genuine improvement over $\operatorname{li}(x)$ in approximating $\pi(x)$ for large $x$.

Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$, and one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive).

QUESTION: Let $\Theta \in [1/2,1]$ denote the supremum of the real parts of the zeros of the Riemann zeta function. The infimum of all $t > 0$ such that $E(x) = o(x^t)$ is $\Theta$. Likewise, the infimum of all $t > 0$ such that $H(x) = o(x^t)$ is also $\Theta$. Clearly, then, the infimum of all $t > 0$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = o(x^t)$ is less than or equal to $\Theta$. Does equality hold? If not, that answers my question fully. If so, let's assume that equality holds. Let $\Theta_1$ be the infimum of all $t \in \mathbb{R}$ such that $E(x) = o(x^\Theta (\log x)^t)$. Then $\Theta_1$ is also the infimum of all $t \in \mathbb{R}$ such that $H(x) = o(x^\Theta (\log x)^t)$. Thus the infimum all $t \in \mathbb{R}$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is $o(x^\Theta (\log x)^t)$ is less than or equal to $\Theta_1$. Does equality hold?

MOTIVATION: Note that $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$. (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li(y)}}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ were smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth (in the sense asked in the QUESTION above), then that would be interesting. It would show that the product $p_{\operatorname{li(x)}}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.

Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. The spirit of this post is to inquire whether or not $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is a genuine improvement over $\operatorname{li}(x)$ in approximating $\pi(x)$ for large $x$.

Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where the functions $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$ have the same "order of growth," while one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive). Thus, the "order of growth" of $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is no larger than that of $\pi(x)-\operatorname{li}(x)$. My question is, is the opposite also true, and thus $\pi(x)-\operatorname{li}(x)$ and $\pi(x) -\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ have the "same order of growth"?

By the "same order of growth" here, I don't mean in terms of $\asymp$ or $\sim$, I mean being $o$ of the same, or almost all of the same, logarithmico-exponential functions (in Hardy's sense).
EDIT: It would take a while to explain exactly what I mean here by "same order of growth," so let me ask a narrower but more precise question.

PRECISE QUESTION: Let $\Theta \in [1/2,1]$ denote the supremum of the real parts of the zeros of the Riemann zeta function. The infimum of all $t > 0$ such that $\pi(x)-\operatorname{li}(x) = o(x^t)$ is $\Theta$. Likewise, the infimum of all $t > 0$ such that $\frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x) = o(x^t)$ is also $\Theta$. Clearly, then, the infimum of all $t > 0$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = o(x^t)$ is less than or equal to $\Theta$. Does equality hold? If so, then consider the infimum of all $t > 0$ such that the three respective functions is $o(x^\Theta (\log x)^t)$, and ask the same questions.

EDIT: Someone asks, what's the point of asking this question? One has $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$, and thus $x \geq p_{\operatorname{li(x)}}$ if and only if $\pi(x) \geq \operatorname{li}(x)$.  (This is a monotone Galois connection from $\mathbb{R}_{>0}$ to itself.) The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ were smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth, then that would be interesting.

Let $p_n$ denote the $n$th prime, let $p_x = p_{\lceil x \rceil}$ for all $x > 0$, let $\pi(x)$ denote the prime counting function, and let $\operatorname{li}(x) = \int_0^x \frac{dt}{\log t}$ denote the logarithmic integral function.

The function $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ seems to approximate $\pi(x)$ much better than $\operatorname{li}(x)$ does, at least for small $x$. The spirit of this post is to inquire whether or not $\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is a genuine improvement over $\operatorname{li}(x)$ in approximating $\pi(x)$ for large $x$.

Note that $$\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = E(x)+H(x),$$ where the functions $E(x) = \pi(x)-\operatorname{li}(x)$ and $H(x) = \frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x)\sim \frac{1}{\log(x)}(p_{\operatorname{li}(x)}-x)$ have the same "order of growth," while one has $E(x)\geq 0$ if and only if $H(x) \leq 0$ (so all "interference" in $E(x)+H(x)$ is destructive). Thus, the "order of growth" of $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ is no larger than that of $\pi(x)-\operatorname{li}(x)$. My question is, is the opposite also true, and thus $\pi(x)-\operatorname{li}(x)$ and $\pi(x) -\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x)$ have the "same order of growth"?

By the "same order of growth" here, I don't mean in terms of $\asymp$ or $\sim$, I mean being $o$ of the same, or almost all of the same, logarithmico-exponential functions (in Hardy's sense).
EDIT: It would take a while to explain exactly what I mean here by "same order of growth," so let me ask a narrower but more precise question.

PRECISE QUESTION: Let $\Theta \in [1/2,1]$ denote the supremum of the real parts of the zeros of the Riemann zeta function. The infimum of all $t > 0$ such that $\pi(x)-\operatorname{li}(x) = o(x^t)$ is $\Theta$. Likewise, the infimum of all $t > 0$ such that $\frac{\operatorname{li}(x)}{p_{\operatorname{li}(x)}}(p_{\operatorname{li}(x)}-x) = o(x^t)$ is also $\Theta$. Clearly, then, the infimum of all $t > 0$ such that $\pi(x)-\frac{x}{p_{\operatorname{li}(x)}}\operatorname{li}(x) = o(x^t)$ is less than or equal to $\Theta$. Does equality hold? If so, then consider the infimum of all $t > 0$ such that the three respective functions is $o(x^\Theta (\log x)^t)$, and ask the same questions.

EDIT: Someone asks, what's the point of asking this question? One has $x \geq p_y$ if and only if $\pi(x) \geq y$, for all $x, y > 0$.   (This defines a monotone Galois connection from $\mathbb{R}_{>0}$ to itself, which expresses an adjoint relationship in the corresponding poset category). Thus $x \geq p_{\operatorname{li(y)}}$ if and only if $\pi(x) \geq \operatorname{li}(y)$. The quantity $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ is thus natural to consider---discrete on one side, continuous on the other, and involving "both sides" of the duality, not just one. If $p_{\operatorname{li(x)}}\pi(x) - x \operatorname{li}(x)$ were smaller than $x(\pi(x)-\operatorname{li}(x))$ in order of growth, then that would be interesting. It would show that the product $p_{\operatorname{li(x)}}\pi(x)$ of the right-left adjoint pair is better approximated by $x \operatorname{li(x)}$ than either adjoint is by $x$ and $\operatorname{li(x)}$, respectively.